Let $u\in L^\infty (\mathbb R^d)$. Is it true that $$\|u^{a}\|_{L^\infty (\mathbb R^d)}=\|u\|_{L^\infty (\mathbb R^d)}^a$$ where $a>0$ ?
Attempt
I would say yes since if $\ell=\|u\|_{L^\infty (\mathbb R^d)}$ then I thing that $\ell^\alpha =\|u^\alpha \|_{L^\infty }$, but I would like a confirmation.
We can find a set $N$ of measure zero such that $|u(x)|\leq\|u\|_{L^{\infty}}$ for all $x\in{\bf{R}}^{d}-N$, for such an $x$, $|u(x)|^{a}\leq\|u\|_{L^{\infty}}^{a}$, so $\|u^{a}\|_{L^{\infty}}\leq\|u\|_{L^{\infty}}^{a}$.
So $\|u\|_{L^{\infty}}=\|(u^{a})^{1/a}\|_{L^{\infty}}\leq\|u^{a}\|_{L^{\infty}}^{1/a}$, then $\|u\|_{L^{\infty}}^{a}\leq\|u^{a}\|_{L^{\infty}}$, so the result follows.
